Let $A, B$ be square matrices of equal size ($n \times n$) and $A^2 + B =A^2B$. Prove that $AB=BA$

Question

I need help with solving the problem about square matrices of equal size. I know that if $A + B = AB$, then $AB = BA$, but I can't prove this one. Please advise how to solve or think about this. Thanks in advance )

Multiply on the left and right by A. Why is that allowed? – Paul Feb 10 '20 at 10:47 — Paul, Feb 10 '20 at 10:47
Sorry for uncertainty, A and B both have size $n*$n – Tigran Petrosyan Feb 10 '20 at 10:56 — Tigran Petrosyan, Feb 10 '20 at 10:56

orangeskid · Accepted Answer · 2020-02-19T01:53:49.790

8

Your equality is equivalent to $$(I-A^2)(I-B)=I$$ so $(I-B)= (I-A^2)^{-1}$. It is enough to check that $A$ commutes with $(I-B)$. But since $A$ commutes with $(I-A^2)$, it will also commute with its inverse.

edited Feb 19 '20 at 01:53

answered Feb 10 '20 at 10:58

orangeskid

53,909

Thanks for help, but here is the proof of last part of your proof ..
https://math.stackexchange.com/questions/1879244/if-an-invertible-matrix-a-commutes-with-c-show-that-a-1-commutes-with
– Tigran Petrosyan Feb 10 '20 at 11:06
@Tigran Petrosyan: Great point, that completes the proof, thanks! – orangeskid Feb 10 '20 at 11:10

score 5 · Answer 2 · answered Feb 10 '20 at 11:00

Left-inverses are right-inverses. In your original problem, $$(I-A)(I-B) = I - A - B + AB = I$$ and so $(I-A)(I-B) = I = (I-B)(I-A)$ giving the result.

In your next example $$I=(I-A^2)(I-B)=(I+A)(I-A)(I-B) \\ \text{and} \\ I = (I-B)(I-A^2)= (I-B)(I-A)(I+A)$$ This gives two expressions for $(I+A)^{-1}$ and the result follows.

Let $A, B$ be square matrices of equal size ($n \times n$) and $A^2 + B =A^2B$. Prove that $AB=BA$

2 Answers2